Interference of light, coherent sources and Young's double slit experiment

Interference and coherent sources

Dip a wire loop in soap solution and look at the film in the light of a bulb: you see bright colours. They come from interference: the redistribution of light energy when waves from two coherent sources superpose. Coherent sources give out waves of the same frequency with a constant phase difference. Two separate bulbs are never coherent. Each sends out light whose phase changes at random, many millions of times a second. Their phase difference keeps changing, so the screen looks evenly lit. So we make coherent sources by dividing light from one source into two parts. In Young's experiment, light from a narrow slit S falls on two narrow slits S₁ and S₂, close together and equally far from S. They pick up parts of the same wavefront, so they act as coherent sources, and the waves beyond them form alternate bright and dark fringes.

Derivation: intensity and the conditions for bright and dark fringes

1. 1
   At a point P, the waves from S₁ and S₂ are y₁ = a sin ωt and y₂ = a sin(ωt + δ), where δ is their phase difference.
2. 2
   By superposition, y = a[sin ωt + sin(ωt + δ)] = 2a cos(δ/2) sin(ωt + δ/2).
3. 3★
   So the resultant amplitude is A = 2a cos(δ/2), and the intensity is I ∝ 4a² cos²(δ/2).
4. 4
   Bright: δ = 0, 2π, 4π ... gives I = 4a², four times the intensity of one wave.
5. 5
   Dark: δ = π, 3π, 5π ... gives I = 0.
6. 6
   A path difference Δ gives a phase difference δ = (2π/λ)Δ. So the point is bright when Δ = nλ and dark when Δ = (2n + 1)λ/2.
7. 7
   So, energy is not destroyed at the dark fringes: it moves to the bright fringes, and the average intensity stays 2a².

★ key step

Derivation: fringe width

The slits S₁ and S₂ are a distance d apart, and the screen is a distance D away, with D much larger than d. A point P on the screen is at a distance x from the centre O.

1. 1
   The path difference at P is Δ = S₂P − S₁P = d sin θ, where θ is the angle of P seen from the slits.
2. 2
   For small θ, sin θ ≈ tan θ = x/D. So Δ = xd/D.
3. 3
   Bright fringe: xd/D = nλ, so xₙ = nλD/d, with n = 0, 1, 2 ...
4. 4
   Dark fringe: xd/D = (2n + 1)λ/2, so xₙ = (n + ½)λD/d.
5. 5★
   The fringe width is the distance between two neighbouring bright fringes: β = xₙ₊₁ − xₙ = λD/d. Dark fringes give the same width.
6. 6
   So, all fringes are equally wide, and they are wider for longer λ or larger D, and narrower for larger d.

★ key step

How to draw the Young's double slit set-up

1. 1
   On the left, draw a source and a narrow slit S.
2. 2
   To its right, draw a barrier with two slits S₁ and S₂, one above and one below the axis. Mark their separation d.
3. 3
   On the far right, draw the screen. Mark the distance D from the slits to the screen.
4. 4
   Draw arcs (wavefronts) spreading from S, and overlapping arcs from S₁ and S₂.
5. 5
   Mark the centre O on the screen and a point P at height x. Join S₁P and S₂P.
6. 6
   On the screen, draw alternate bright and dark bands. Label central bright fringe at O and β between two bright fringes.
7. 7
   Beside the screen, sketch the intensity: equal peaks at the bright fringes and zeros at the dark fringes, with a dotted line at the average level.

Worked example: fringe width and fringe positions

How the pattern changes

• β ∝ λ: red light gives wider fringes than blue light.
• β ∝ D: moving the screen away from the slits widens the fringes.
• β ∝ 1/d: bringing the slits closer together widens the fringes.
• In water (μ = 4/3), λ becomes λ/μ, so β becomes 3/4 of its value in air.
• One slit closed: the fringes disappear, and the screen shows only the spread-out light from one slit.
• White light: the central fringe is white, with a few coloured fringes on each side, and then uniform light.
• Wider slits: the fringes become less sharp, and they fade when the slits are too wide.